GSEB Class 11 Maths Solutions Chapter 15 Statistics Exercise 15.3

Question 1. From the data given below, state which group is more variable A or B?

Answer:
For Group A:
\( \bar{x} = A + \frac{\sum f_i y_i}{\sum f_i} \times h = 45 + \frac{-6}{150} \times 10 = 45 - 0.4 = 44.6 \)
\( \sigma^2 = \frac{h^2}{N^2} [N \sum f_i y_i^2 - (\sum f_i y_i)^2] \)
\( \implies \) \( \sigma^2 = \frac{100}{22500} [150 \times 342 - (-6)^2] \)
\( \implies \) \( \sigma^2 = \frac{1}{225} [51300 - 36] = \frac{51264}{225} = 227.84 \)
\( \sigma = \sqrt{227.84} = 15.09 \)
Coefficient of variance (C.V.) \( = \frac{\sigma}{\bar{x}} \times 100 = \frac{15.09}{44.6} \times 100 = 33.83 \)
For Group B:
\( \bar{x} = A + \frac{\sum f_i y_i}{\sum f_i} \times h = 45 + \frac{-6}{150} \times 10 = 45 - 0.4 = 44.6 \)
\( \sigma^2 = \frac{h^2}{N^2} [N \sum f_i y_i^2 - (\sum f_i y_i)^2] \)
\( \implies \) \( \sigma^2 = \frac{100}{22500} [150 \times 592 - (-6)^2] \)
\( \implies \) \( \sigma^2 = \frac{1}{225} [88800 - 36] = \frac{88764}{225} = 394.50 \)
\( \sigma = \sqrt{394.50} = 19.86 \)
Coefficient of variance (C.V.) \( = \frac{\sigma}{\bar{x}} \times 100 = \frac{19.86}{44.6} \times 100 = 44.53 \)
The coefficient of variation in group B is greater than the coefficient of variation in group A. This means group B is more variable than group A.
In simple words: We calculate how spread out the data is for each group. Group B has a higher "coefficient of variation" score, which tells us its marks are more varied compared to Group A.

Exam Tip: Remember that a higher coefficient of variation indicates greater variability, while a lower one suggests more consistency or less spread in the data.

 

Question 2. From the prices of shares X and Y below, find out which is more stable in value:

Answer:
For shares X:
Mean \( \bar{x} = A + \frac{\sum y_i}{N} = 52 + \frac{-10}{10} = 51 \)
Standard Deviation \( \sigma = \frac{1}{N} \sqrt{N \sum y_i^2 - (\sum y_i)^2} \)
\( \implies \) \( \sigma = \frac{1}{10} \sqrt{10 \times 360 - (-10)^2} \)
\( \implies \) \( \sigma = \frac{1}{10} \sqrt{3600 - 100} = \frac{\sqrt{3500}}{10} = \frac{59.16}{10} = 5.916 \)
Coefficient of variation \( = \frac{\sigma}{\bar{x}} \times 100 = \frac{5.916}{51} \times 100 = 11.6 \)
For shares Y:
Mean \( \bar{x} = A + \frac{\sum y_i}{N} = 105 + \frac{0}{10} = 105 \)
Standard Deviation \( \sigma = \frac{1}{N} \sqrt{N \sum y_i^2 - (\sum y_i)^2} \)
\( \implies \) \( \sigma = \frac{1}{10} \sqrt{10 \times 40 - 0^2} \)
\( \implies \) \( \sigma = \frac{1}{10} \sqrt{400 - 0} = \frac{20}{10} = 2 \)
Coefficient of variation \( = \frac{\sigma}{\bar{x}} \times 100 = \frac{2}{105} \times 100 = 1.9 \)
The coefficient of variation for shares Y is less than that for shares X. Therefore, share Y is more stable than share X.
In simple words: To find which share is more steady, we compare their 'coefficient of variation'. Share Y has a smaller coefficient of variation, meaning its prices change less and are more stable than share X.

Exam Tip: Stability in value is indicated by a lower coefficient of variation. Always clearly state your conclusion based on the calculated values.

 

Question 3. An analysis of monthly wages paid to workers in two firms A and B belonging to the same industry, give the following results:

(i) Which firm A or B pays out larger amount of monthly wages?
(ii) Which firm A or B, shows greater variability in individual wages?
Answer:
For firm A:
Number of wage earners = 586
Mean of monthly wages \( \bar{x} \) = Rs 5253
Amount paid by firm A = Rs \( (586 \times 5253) \) = Rs 3078258
Variance of distribution of wages = 100
Standard deviation \( \sigma = \sqrt{\text{Variance}} = \sqrt{100} = 10 \)
Coefficient of variation \( = \frac{\sigma}{\bar{x}} \times 100 = \frac{10}{5253} \times 100 = 0.19 \)
For firm B:
Number of wage earners = 648
Mean of monthly wages \( \bar{x} \) = Rs 5253
Amount paid by firm B = Rs \( (648 \times 5253) \) = Rs 3403944
Variance of distribution of wages = 121
Standard deviation \( \sigma = \sqrt{\text{Variance}} = \sqrt{121} = 11 \)
Coefficient of variation \( = \frac{\sigma}{\bar{x}} \times 100 = \frac{11}{5253} \times 100 = 0.21 \)
(i) Monthly wages paid by firm A = Rs 3078258
Monthly wages paid by firm B = Rs 3403944
Firm B pays out a larger amount as monthly wages.
(ii) Coefficient of variation of wages of firm A = 0.19
Coefficient of variation of wages of firm B = 0.21
Since the coefficient of variation for firm B is greater, firm B shows greater variability in individual wages.
In simple words: First, we find the total money each firm pays. Firm B pays more overall. Then, we look at how much the individual wages change. Firm B's wages show more variation, meaning they are less consistent among workers.

Exam Tip: When comparing total amounts, multiply the number of earners by the mean wage. For variability, always compare the coefficients of variation.

 

Question 4. The following is the record of goals scored by team A in a football session: For the team B, mean number of goals scored per match was 2 with standard deviation 1.25 goals. Find which team may be considered more consistent?

Answer:
For Team A:
Mean \( \bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{50}{25} = 2 \)
Standard deviation \( \sigma = \frac{1}{N} \sqrt{N \sum f_i x_i^2 - (\sum f_i x_i)^2} \)
\( \implies \) \( \sigma = \frac{1}{25} \sqrt{25 \times 130 - (50)^2} \)
\( \implies \) \( \sigma = \frac{1}{25} \sqrt{3250 - 2500} = \frac{\sqrt{750}}{25} = \frac{27.386}{25} = 1.095 \)
Coefficient of variation \( = \frac{\sigma}{\bar{x}} \times 100 = \frac{1.095}{2} \times 100 = 54.75 \)
For Team B:
Mean \( \bar{x} \) = 2
Standard Deviation \( \sigma \) = 1.25
Coefficient of variation \( = \frac{\sigma}{\bar{x}} \times 100 = \frac{1.25}{2} \times 100 = 62.5 \)
The coefficient of variation of goals for team A is less than that for team B. Therefore, team A is more consistent than team B.
In simple words: We calculate the "consistency" for each team. Team A's coefficient of variation is lower, which means their goal scoring is more steady and predictable compared to Team B.

Exam Tip: Consistency is the opposite of variability; a lower coefficient of variation indicates greater consistency. Make sure to clearly state which team is more consistent based on this comparison.

 

Question 5. The sum and sum of squares corresponding to length x (in cm) and weight y (in grams) of 50 plant products are given below: \( \sum_{i=1}^{50}x_i = 212 \), \( \sum_{i=1}^{50}x_i^2 = 902.8 \), \( \sum_{i=1}^{50}y_i = 261 \), \( \sum_{i=1}^{50}y_i^2 = 1457.6 \). Which is more varying, the length or weight?
Answer:
Given: \( N = 50 \)
For Length (x):
\( \sum x_i = 212 \)
\( \sum x_i^2 = 902.8 \)
Mean \( \bar{x} = \frac{\sum x_i}{N} = \frac{212}{50} = 4.24 \text{ cm} \)
Standard deviation \( \sigma = \frac{1}{N} \sqrt{N \sum x_i^2 - (\sum x_i)^2} \)
\( \implies \) \( \sigma = \frac{1}{50} \sqrt{50 \times 902.8 - (212)^2} \)
\( \implies \) \( \sigma = \frac{1}{50} \sqrt{45140 - 44944} = \frac{\sqrt{196}}{50} = \frac{14}{50} = 0.28 \)
Coefficient of variation (C.V.) \( = \frac{\sigma}{\bar{x}} \times 100 = \frac{0.28}{4.24} \times 100 = 6.6 \)
For Weight (y):
\( \sum y_i = 261 \)
\( \sum y_i^2 = 1457.6 \)
Mean \( \bar{x} = \frac{\sum y_i}{N} = \frac{261}{50} = 5.22 \text{ g} \)
Standard deviation \( \sigma = \frac{1}{N} \sqrt{N \sum y_i^2 - (\sum y_i)^2} \)
\( \implies \) \( \sigma = \frac{1}{50} \sqrt{50 \times 1457.6 - (261)^2} \)
\( \implies \) \( \sigma = \frac{1}{50} \sqrt{72880 - 68121} = \frac{\sqrt{4759}}{50} = \frac{68.9855}{50} = 1.38 \)
Coefficient of variation (C.V.) \( = \frac{\sigma}{\bar{x}} \times 100 = \frac{1.38}{5.22} \times 100 = 26.4 \)
The coefficient of variation of weight (26.4) is more than that of length (6.6). Therefore, weight is more varying than length.
In simple words: To see which measurement, length or weight, changes more, we compare their 'coefficient of variation' scores. Weight has a much higher score, meaning it varies a lot more compared to length in these plant products.

Exam Tip: When comparing variability between two different types of measurements (like length and weight), always use the Coefficient of Variation as it accounts for differences in their respective means.